dfs(nums) that returns the number of valid permutations for a given subtree.nums has less than 3 elements, return 1 as there is only one way to form the BST.nums as the root and partition the rest into left_nodes and right_nodes based on their values relative to the root.dfs on left_nodes and right_nodes to get the number of valid permutations for each subtree.C(left+right, left) to determine the number of ways to interleave the two subtrees.nums.dfs(nums) modulo 10^9 + 7.